Reputation: 4854
fn main() {
let arr: [u8;8] = [97, 112, 112, 108, 101];
println!("Len is {}",arr.len());
println!("Elements are {:?}",arr);
}
error[E0308]: mismatched types
--> src/main.rs:2:23
|
2 | let arr: [u8;8] = [97, 112, 112, 108, 101];
| ------ ^^^^^^^^^^^^^^^^^^^^^^^^ expected an array with a fixed size of 8 elements, found one with 5 elements
| |
| expected due to this
Is there any way to pad the remaining elements with 0's? Something like:
let arr: [u8;8] = [97, 112, 112, 108, 101].something();
Upvotes: 11
Views: 7494
Reputation: 2654
In addition to the other answers, you can use const generics to write a dedicated method.
fn pad_zeroes<const A: usize, const B: usize>(arr: [u8; A]) -> [u8; B] {
assert!(B >= A); //just for a nicer error message, adding #[track_caller] to the function may also be desirable
let mut b = [0; B];
b[..A].copy_from_slice(&arr);
b
}
Upvotes: 10
Reputation: 1303
You could start with zeros and set the initial values afterwards. This requires arr
to be mut
though.
fn main() {
let mut arr: [u8;8] = [0;8];
let init = [97, 112, 112, 108, 101];
arr[..init.len()].copy_from_slice(&init);
println!("Len is {}",arr.len());
println!("Elements are {:?}",arr);
}
Link to Playground
Upvotes: 5
Reputation: 2592
You can use concat_arrays macro for it:
use concat_arrays::concat_arrays;
fn main() {
let arr: [u8; 8] = concat_arrays!([97, 112, 112, 108, 101], [0; 3]);
println!("{:?}", arr);
}
I don't think it's possible to do without external dependencies.
Upvotes: 10